 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

A person trying to lose weight by burning fat lifts a mass of $10$ $kg$ upto a height of $1$ $m$ $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times {10^7}J$ of energy per $kg$ which is converted to mechanical energy with a $20\%$ efficiency rate. Take $g = 9.8\,m{s^{ - 2}}$ :
A
$9.89 \times {10^{ - 3}}\,\,kg$
B
$12.89 \times {10^{ - 3}}\,kg$
C
$2.45 \times {10^{ - 3}}\,\,kg$
D
$6.45 \times {10^{ - 3}}\,\,kg$

Explanation

Assume the amount of fat is used = x kg

So total Mechanical energy available through fat

= $x \times 3.8 \times {10^7} \times {{20} \over {100}}$

And work done through lifting up

= 10 $\times$ 9.8 $\times$ 1000 = 98000 J

$\Rightarrow$ $x \times 3.8 \times {10^7} \times {{20} \over {100}}$ = 98000

$\Rightarrow$ $x$ = 12.89 $\times$ 10-3 kg
2

JEE Main 2014 (Offline)

When a rubber-band is stretched by a distance $x$, it exerts restoring force of magnitude $F = ax + b{x^2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is :
A
$a{L^2} + b{L^3}$
B
${1 \over 2}\left( {a{L^2} + b{L^3}} \right)$
C
${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$
D
${1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)$

Explanation

Given Restoring force, F = ax + bx2

Work done in stretching the rubber-band by a distance $dx$ is

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$

Intergrating both sides,

$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$

= $\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$

= ${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$
3

AIEEE 2012

This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs ${S_1}$ and ${S_2}$ of force constants ${k_1}$ and ${k_2}$, respectively, are stretched by the same force, it is found that more work is done on spring ${S_1}$ than on spring ${S_2}$.

STATEMENT 1: If stretched by the same amount work done on ${S_1}$, Work done on ${S_1}$ is more than ${S_2}$
STATEMENT 2: ${k_1} < {k_2}$

A
Statement 1 is false, Statement 2 is true
B
Statement 1 is true, Statement 2 is false
C
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
D
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

Explanation

We know force (F) = kx

$W = {1 \over 2}k{x^2}$

$W =$ ${{{{\left( {kx} \right)}^2}} \over {2k}}$ $\,\,\,$

$\therefore$ $W = {{{F^2}} \over {2k}}$ [ as $F=kx$ ]

When force is same then,

$W \propto {1 \over k}$

Given that, ${W_1} > {W_2}$

$\therefore$ ${k_1} < {k_2}$

Statement-2 is true.

For the same extension, x1 = x2 = x

Work done on spring S1 is W1 = ${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$

Work done on spring S2 is W2 = ${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$

$\therefore$ ${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$

As ${k_1} < {k_2}$ then ${W_1} < {W_2}$

So, Statement-1 is false.
4

AIEEE 2010

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$ where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$ is
A
${{{b^2}} \over {2a}}$
B
${{{b^2}} \over {12a}}$
C
${{{b^2}} \over {4a}}$
D
${{{b^2}} \over {6a}}$

Explanation

Given $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$

${U\left( {x = \infty } \right)}$ = 0

We know $F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$

At equilibrium: ${{dU\left( x \right)} \over {dx}} = 0$

$\Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}}$

$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$

$\therefore$ ${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$

$= - {{{b^2}} \over {4a}}$

$\therefore$ $D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$